Question

f.

$$ \\lim _ { x \ ightarrow 0 } \\frac { x - \\sin x } { x ^ { 3 } + x ^ { 2 } } $$

g.

$$ \\lim _ { x \ ightarrow 0 } ( 1 + 2 x ) ^ { 1 / x } $$

h.

$$ \\lim _ { x \ ightarrow 0 ^ { + } } x ^ { 2 } \\ln ( x ) $$

i.

$$ \\lim _ { x \ ightarrow \\infty } \\left( x ^ { 2 } + 1 \ ight) ^ { 1 / \\ln x } $$

j.

$$ \\lim _ { x \ ightarrow 0 ^ { + } } x ^ { x } $$

Explanation:

Step1: Apply L'Hospital's Rule for F

For $\lim_{x
ightarrow0}\frac{x - \sin x}{x^{3}+x^{2}}$, it is in $\frac{0}{0}$ form.
Differentiate numerator and denominator:

$$\lim_{x ightarrow0}\frac{1-\cos x}{3x^{2}+2x}$$

Still $\frac{0}{0}$ form. Differentiate again:

$$\lim_{x ightarrow0}\frac{\sin x}{6x + 2}=\frac{0}{2}=0$$

Step2: Use the formula for G

For $\lim_{x
ightarrow0}(1 + 2x)^{1/x}$, let $y=(1 + 2x)^{1/x}$. Then $\ln y=\frac{\ln(1 + 2x)}{x}$.
Apply L'Hospital's Rule (since $\frac{0}{0}$ form):

$$\lim_{x ightarrow0}\frac{\frac{2}{1+2x}}{1}=2$$

So $\lim_{x
ightarrow0}y = e^{2}$

Step3: Rewrite for H

For $\lim_{x
ightarrow0^{+}}x^{2}\ln x$, rewrite as $\lim_{x
ightarrow0^{+}}\frac{\ln x}{x^{-2}}$. It is in $\frac{-\infty}{\infty}$ form.
Apply L'Hospital's Rule:

$$\lim_{x ightarrow0^{+}}\frac{\frac{1}{x}}{-2x^{-3}}=\lim_{x ightarrow0^{+}}\frac{-x^{2}}{2}=0$$

Step4: Take natural log for I

Let $y=(x^{2}+1)^{1/\ln x}$. Then $\ln y=\frac{\ln(x^{2}+1)}{\ln x}$.
Apply L'Hospital's Rule (as $x
ightarrow\infty$, $\frac{\infty}{\infty}$ form):

$$\lim_{x ightarrow\infty}\frac{\frac{2x}{x^{2}+1}}{\frac{1}{x}}=\lim_{x ightarrow\infty}\frac{2x^{2}}{x^{2}+1}=2$$

So $\lim_{x
ightarrow\infty}y = e^{2}$

Step5: Rewrite for J

Let $y = x^{x}$. Then $\ln y=x\ln x=\frac{\ln x}{x^{-1}}$. As $x
ightarrow0^{+}$, $\frac{-\infty}{\infty}$ form.
Apply L'Hospital's Rule:

$$\lim_{x ightarrow0^{+}}\frac{\frac{1}{x}}{-x^{-2}}=\lim_{x ightarrow0^{+}}(-x)=0$$

So $\lim_{x
ightarrow0^{+}}y = e^{0}=1$

Answer:

F. $0$
G. $e^{2}$
H. $0$
I. $e^{2}$
J. $1$